Suppose $R$ is a commutative ring, $a, b$ are zero divisors of $R$ such that $ab$ is nonzero. Show that $ab$ is also a zero divisor of $R$.

Question

I am going to show that

Suppose $R$ is a commutative ring, $a, b$ are zero divisors of $R$ such that $ab$ is nonzero. Show that $ab$ is also a zero divisor of $R$.

To show that $ab$ is a zero divisor of R, we proceed to find a nonzero element $x\in R$ such that $(ab)x = 0$. Since $a$ and $b$ are zero divisors, there exist nonzero elements $r$, $s$ ∈ R such that $ar = 0$ and $bs = 0$. Now, consider the element $x = rs$. Note that $x\not= 0$ because suppose NOT, i.e. $x = 0$, then $rs = 0$, which would contradict the fact that $a$ and $b$ are zero divisors (We have $0 = a(rs) = (ar)s$ or $0 = b(rs) = (bs)r$ (since $R$ is commutative) and one of $r$ or $s$ would have to be zero).

Let’s compute $(ab)x$:

$(ab)x = (ab)(rs) = a(b(rs)) = a((bs)r) = a(0 \cdot r) = a \cdot 0 = 0$

Therefore, $ab$ is a zero divisor of $R$.

Sorry my proof looks clumsy and not sure it is correct or not. Can anyone kindly help me to have a look? Thanks.

Answer 1

You really only need to use $x = s$, as $$ (ab)s = a(bs) = a\cdot 0 = 0 $$ It's a lot easier to show that $x\neq 0$ in this case. It isn't actually possible to show in general for $x = rs$. Consider, for instance, $\Bbb Z^3$ with $a = b = (1,0,0)$, $r = (0,1,0)$ and $s = (0,0,1)$. This means you can't just pick arbitrary $r$ and $s$, but instead be careful.

Answer 2

You should prove the more general

Suppose $R$ is a commutative ring, $a$ is a zero divisor of $R$ and $b$ an element of $R$ such that $ab \ne 0$. Then $ab$ is a zero divisor of $R$.

Since $a$ is a zero divisor, there exists $x \in R \setminus \{0\}$ such that $ax = 0$. But then $$(ab)x = (ba)x = b(ax) = b0 = 0 .$$